Jovian planets

What dna is kept in the nucleus and is enclosed by its own membrane

lyudmila [28]

In this, the DNA is kept in the nucleus, which is enclosed by its own membrane. An organelle in a plant cell that uses the energy from sunlight to make sugar. When a cell and its organelles are organized to allow it to do a specific job. An organelle that releases stored energy, which comes mostly from sugars.

3 0

3 years ago

If a single bacteriophage infects one

melomori [17]

The answer is 3,625,908,203,125. This calculated by (325)^5 .

This is exponential growth where the population of the virus doubles in every lytic cycle and decimates the E. coli population. In some hours the lawn of E. coli cells will be completely mowed down. The other life cycle of bacteriophages, other than lytic, is lysogenic cycle that does not lyse the bacteria but only integrates its DNA in the genome and is replicated when the bacteria divides.

5 0

3 years ago

What is the acceleration of a 50 kg object pushed with a force of 500 newtons

tatiyna

Force is equal to mass times acceleration, therefore acc. is equal to Force divided by mass.

500 N /50kg equals 10m/s^2.

3 0

4 years ago

In Drosophila melanogaster, vestigial wings (vg) is recessive to normal wings (vg+), black body (b) is recessive to gray body (b

yan [13]

Correct progeny phenotype:

1779 vestigial wings, black body and purple eyes, vg b pr
1665 normal wings, a grey body and red eyes, vg+ b+ pr+
252 normal wings, an black body and purple eyes, vg+ b pr
241 vestigial wings, a gray body and red eyes, vg b+ pr+
131 normal wings, an black body and red eyes, vg +b pr+
118 vestigial wings, a gray body and purple eyes, vg b+ pr
13 vestigial wings, an black body and red eyes, vg b pr+
9 normal wings, a gray body and purple eyes, vg+ b+ pr

Answer:

The order of these genes is vg --- pr --- b
Map distances between the genes vg/pr = 12.2 MU
Map distance between the genes pr/b = 6.4 MU
Map distances between the genes vg/b = 18.6 MU

Explanation:

We know that

•Normal wings expressed by vg+ is dominant over vestigial wings, vg

•Gray body b+ is dominant over black body

•Red eyes, pr+, is dominant over purple ayes, pr

We have the number of descendants of each phenotype product of the tri-hybrid cross.

•1779 vestigial wings, black body and purple eyes vg b pr

•1665 normal wings, a grey body and red eyes vg+ b+ pr+

•252 normal wings, an black body and purple eyes vg+ b pr

•241 vestigial wings, a gray body and red eyes vg b+ pr+

•131 normal wings, an black body and red eyes vg +b pr+

• 118 vestigial wings, a gray body and purple eyes vg b+ pr

•13 vestigial wings, an black body and red eyes vg b pr+

• 9 normal wings, a gray body and purple eyes vg+ b+ pr

The total number of individuals is 4208.

In a tri-hybrid cross, it can occur that the three genes assort independently or that two of them are linked and the third not, or that the three genes are linked. In this example, in particular, the three genes are linked on the same chromosome.

Knowing that the genes are linked, we can calculate genetic distances between them. First, we need to know their order in the chromosome, and to do so, we need to compare the genotypes of the parental gametes with the ones of the double recombinants. We can recognize the parental gametes in the descendants because their phenotypes are the most frequent, while the double recombinants are the less frequent. So:

Parental)

1779 vestigial wings, black body and purple eyes vg b pr

1665 normal wings, a grey body and red eyes vg+ b+ pr+

Double recombinant)

13 vestigial wings, an black body and red eyes vg b pr+
9 normal wings, a gray body and purple eyes vg+ b+ pr

Simple recombinant)

252 normal wings, an black body and purple eyes vg+ b pr
241 vestigial wings, a gray body and red eyes vg b+ pr+
131 normal wings, an black body and red eyes vg +b pr+
118 vestigial wings, a gray body and purple eyes vg b+ pr

Comparing parental with the double recombinants we will realize that between

vg b pr (parental)

vg b pr+ (double recombinant)

and

vg+ b+ pr+ (Parental)

vg+ b+ pr (double recombinant)

They only change in the position of the alleles pr/pr+. This suggests that the position of the gene pr is in the middle of the other two genes, vg and b, because in a double recombinant only the central gene changes position in the chromatid.

So, the order of the genes is:

---- vg ---- pr -----b ----

Now we will call Region I to the area between vg and pr and Region II to the area between pr and b.

Once established the order of the genes we can calculate distances between them, and we will do it from the central gene to the genes on each side. First We will calculate the recombination frequencies, and we will do it by region. We will call P1 to the recombination frequency between vg and pr genes, and P2 to the recombination frequency between pr and b.

P1 = (R + DR) / N

P2 = (R + DR)/ N

Where: R is the number of simple recombinants in each region, DR is the number of double recombinants in each region, and N is the total number of individuals.

So:

Parental)

• 1779 vestigial wings, black body and purple eyes vg pr b

• 1665 normal wings, a grey body and red eyes vg+ pr+ b+

Double recombinant)

• 13 vg pr+ b

• 9 vg+ pr b+

Simple recombinant)

• 252 vg+ pr b

• 241 vg pr+ b+

• 131 vg+ pr+ b

• 118 vg pr b+

P1 = (R + DR) / N

P1 = (252+241+13+9)/4208

P1 = 515/4208

P1 = 0.122

P2= (R + DR) / N

P2 = (131+118+13+9)/4208

P2 = 271/4208

P2 = 0.064

Now, to calculate the recombination frequency between the two extreme genes, vg and b, we can just perform addition or a sum:

P1 + P2= Pt

0.122 + 0.064 = Pt

0.186=Pt

The genetic distance will result from multiplying that frequency by 100 and expressing it in map units (MU). One centiMorgan (cM) equals one map unit (MU).

The map unit is the distance between the pair of genes. Every 100 meiotic products, one of them results in a recombinant product. Now we must multiply each recombination frequency by 100 to get the genetic distance in map units:

GD1= P1 x 100 = 0.122 x 100 = 12.2 MU

GD2= P2 x 100 = 0.064 x 100 = 6.4 MU

GD3=Pt x 100 = 0.186 x 100 = 18.6 MU

---- vg ---------------------- pr ---------------------b ----

R1 R2

-----vg----12.2MU---------pr—

----pr--------6.4 MU----b—

-----vg ----------------18.6 MU--------------------b----

3 0

3 years ago

The client reports to the nurse that she feels as if her eyes are persistently dry. this symptom is consistent with a deficiency

g100num [7]

The symptom that the client is experiencing is consistent with the deficiency in the Vitamin A which is a dietary element. The vitamin A is responsible for causing dryness to the eyes if a person is lacking this type of dietary element as Vitamin A is the one responsible for providing healthy vision for an individual.

7 0

4 years ago